MATH 152 Problem set 1 solutions. 1. Factorize n4 + n2 + 1: n4 + n2 +1 = n4 +
2n2 + 1 − n2. = (n2 + 1)2 − n2. = (n2 + n + 1)(n2 − n + 1). Since both factors are ...
MATH 152 Problem set 1 solutions
1. Factorize n4 + n2 + 1: n4 + n2 + 1 = n4 + 2n2 + 1 − n2 = (n2 + 1)2 − n2 = (n2 + n + 1)(n2 − n + 1).
Since both factors are greater than 1 when n > 1, it follows that it is not prime. √ √ 2. (i) Proof by contradiction: suppose p is rational. Then we can write p = ab , where a, b are positive integers. Squaring both sides, we obtain a2 a 2 . p= 2 = b b This implies that ab is an integer (otherwise, then b - a, so b2 - a2 , and consequently p is not an integer, a contradiction). Therefore p is a square; in particular, it has at least three different divisors. But this contradicts the assumption that p is prime. √ (ii) We prove the contrapositive statement. Suppose n is rational, i.e. we can write √ a n = b for some positive integers a, b. As before, square both sides to get n= Again
a b
a2 a 2 = . b2 b
has to be an integer. Therefore n is a square of an integer.
(iii) Suppose α is not irrational. Then we want to show that α is an integer. To be more precise, we show that α 6∈ Q \ Z. Write α = rs where r, s ∈ Z and (r, s) = 1. Then r n−1 r n + a1 + . . . + an = 0. s s Multiplying both sides by sn : rn + a1 rn−1 s + . . . + an sn = 0. Therefore rn = −(a1 rn−1 s + . . . + an sn ). The right-hand side here is divisible by s, and therefore so is the left-hand side rn . But since (r, s) = 1, (rn , s) = 1. Hence s = ±1 is forced.
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3. Let p be a prime. The strategy is to compare the power of p in the factorization of the numerator with that of the denominator, and show that the former is no less than the latter for any prime p. The power of p in the numerator is ∞ j X 30n k m=1
pm
j n k + m . p
(Here bxc denotes the greatest integer less than or equal to x.) And the power of p in the denominator is ∞ j X 15n k m=1
pm
+
j 10n k pm
+
j 6n k pm
.
c + b pnm c ≥ b 15n c + b 10n c + b p6n It suffices to show that b 30n m c for each m. To simplify pm pm pm notation a little bit, let pnm = N + α, where N = b pnm c and α = pnm − N . Note that here 0 ≤ α < 1. Then 30n = 30N + 30α, which gives b 30n c = b30αc. By the same logic, pm pm 10n 6n 15n b pm c = b15αc, b pm c = b10αc, b pm c = b6αc. So all we really need to show is b30αc + bαc ≥ b15αc + b10αc + b6αc for any 0 ≤ α < 1. We divide the proof into 30 cases: for each a ∈ {0, 1, 2, . . . , 29}, we show that for a/30 ≤ α < (a + 1)/30 the above inequality holds. The verification of this claim is easy and not as tedious as it seems, it is left to the reader. To give an idea, the first few cases go like: Case a = 0: b30αc = b15αc = b10αc = b6αc = 0. Case a = 1: b30αc = 1, and b15αc = b10αc = b6αc = 0. Case a = 2: b30αc = 2, b15αc = 1, b10αc = b6αc = 0. Case a = 3: b30αc = 3, b15αc = 1, b10αc = 1, b6αc = 0. Case a = 4: b30αc = 4, b15αc = 2, b10αc = 1, b6αc = 0. . . . and so on. In general, b15αc increases by 1 as a increases by 2, b10αc increases as a increases by 3, and b6αc increases by 1 as a increases by 5.
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4. Suppose first that a | bc. Write a = a0 (a, b) and b = b0 (a, b). Note that (a0 , b0 ) = 1. Then a0 (a, b) | b0 (a, b)c ⇒ a0 | b0 c, but since (a0 , b0 ) = 1 we have a0 | c. Conversely, if a0 | c then a0 | b0 c so a | bc.
5. Pick 1 ≤ a < b ≤ n. Then (an! + 1, bn! + 1) = (an! + 1, (b − a)n!). But then any prime divisor of an! + 1 is greater than n, whereas any prime divisor of (b − a)n! is no greater than n. Therefore (an! + 1, bn! + 1) = 1. As a consequence, for any n we can find n distinct integers greater than 1 that are pairwise coprime. But if there were no more than, say, N < ∞ primes, then any N + 1 numbers greater than 1 is not pairwise coprime, by the pigeonhole principle. (To elaborate, make N boxes corresponding to the N primes and “put” a number into a box corresponding to any of its prime divisors. If we put N + 1 numbers in, then at least one box must have at least two numbers in it.)
6. We are asked to compute the maximum power of 10 dividing 2010!. To do this we compute the maximum power of 2 and 5 dividing 2010! and take the smaller of the two. The former equals 10 j X 2010 k = 2002 i 2 i=1 and the latter is
4 j X 2010 k i=1
5i
Therefore the answer is 501.
3
= 501.
